Linear Algebra Problems 01
Table of Contents
Quadratic Curve through three points
This problem is the same as solving a linear system of equations. Suppose the curve is $ax^2+bx+c=y$. Then,
$$ \begin{aligned} &a j^2+b j+c=k \\ &a p^2+b p+c=q \\ &a u^2+b u+c=v \end{aligned} $$If there exists a solution. Then we can find such a curve.
Eigenvalues and Eigenvectors of a Special Matrix
Denote this matrix as $\Omega=M+(n-1)I$. Where $M$ is the matrix with all ones, and $I$ is the identity matrix.
Matrix $M$ has eigenvalues of $0$ with multiplicity of $n-1$ and eigenvalue $n$ with multiplicity $1$. To see why this is the case, see the reference. A short derivation is that, the rank of $M$ is 1, which means the number of nonzero is 1. Also not that the trace of the $M$ is $n$, which is the sum of the eigenvalues. Thus, $M$ has only 1 nonzero eigenvalue of $n$, and all the others are just 0.
Knowing this, multiply any vectors $v$ to the equation above, we have $\Omega v = Mv+(n-1)v$, which can be simplified as $\Omega v=(\lambda+(n-1))v$. This is just the definition of eigenvalues of $\Omega$. Use $\lambda=n,0$, we have $\Omega$ eigenvalues of $2n-1$ and $n-1$ (with multiplicity of $n-1$).
